A proof of a Dodecahedron conjecture for distance sets
AA proof of a Dodecahedron conjecture for distance sets
Hiroshi Nozaki and Masashi ShinoharaSeptember 29, 2020
Abstract
A finite subset of a Euclidean space is called an s -distance set if there exist exactly s values of the Euclidean distances between two distinct points in the set. In this paper,we prove that the maximum cardinality among all 5-distance sets in R is 20, and every5-distance set in R with 20 points is similar to the vertex set of a regular dodecahedron. Key words : Distance sets, dodecahedron.
For X ⊂ R d , let A ( X ) = { d ( x, y ) | x, y ∈ X, x (cid:54) = y } , where d ( x, y ) is the Euclidean distance between x and y . We call X an s -distance set if | A ( X ) | = s . Two s -distance sets are said to be isomorphic if there exists a similar transfor-mation from one to the other. One of the major problems in the theory of distance sets isto determine the maximum cardinality g d ( s ) of s -distance sets in R d for given s and d , andclassify distance sets in R d with g d ( s ) points up to isomorphism. An s -distance set X in R d issaid to be optimal if | X | = g d ( s ). Clearly g ( s ) = s + 1, and the optimal s -distance set is theset of s + 1 points on the line whose two consecutive points have an equal interval. For thecases where d = 2 or s = 2, s -distance sets in R d are well studied [1, 2, 8, 9, 12, 13, 16, 17, 20],because of their simple structures or the relationship to graphs, see Table 1. For d ≤
8, themaximum cardinality g d (2) are determined, and optimal 2-distance sets in R d are classifiedexcept for d = 8 [8, 13]. Moreover, it is known that g (3) = 12, g (4) = 13 and g (4) = 16,and the classification is complete for the three cases [18, 19]. In particular, we recall the clas-sification of optimal s -distance sets in R d for ( d, s ) = (2 , , (3 ,
3) and (3 ,
4) as in Theorem 1.1. d g d (2) 5 6 10 16 27 29 45 s g ( s ) 5 7 9 12 13Table 1: Maximum cardinalities of s -distance sets in R d a r X i v : . [ m a t h . M G ] S e p heorem 1.1. ([17, 18, 19])(1) Every -point -distance set in R is isomorphic to the vertices of the regular nonagonor one of the three configurations given in Figure 1 (a)–(c). Moreover, every -point -distance set in R is isomorphic to the vertices of the regular octagon, the vertices of theregular septagon with its center, Figure 1 (d) or -point subsets of a -point -distanceset. (2) Every -point -distance set in R is isomorphic to the vertices of the icosahedron. (3) Every -point -distance set in R is isomorphic to the vertices of the icosahedron withits center point or the vertex set of the cuboctahedron with its center point. (a) (b) (c) (d)Figure 1: Maximal planar 4-distance setsFor a 2-distance set X , we consider the graph on X where two vertices are adjacent ifthey have the smallest distance in X . We can construct the 2-distance set that has thestructure of a given graph [8]. Lisonˇek [13] gave an algorithm for a stepwise augmentation ofrepresentable graphs (adding one vertex per iteration), and classified the optimal 2-distancesets in R d for d ≤ s -distance sets and classified optimal s -distance sets for ( d, s ) = (2 , , (3 , , (4 , s and d . In the present paper, we add geometricalobservations in R to this algorithm, and obtain the main theorem as follows. Theorem 1.2.
Every -point -distance set in R is isomorphic to the vertices of a regulardodecahedron. In particular, g (5) = 20 . This was a long standing open problem [5] as well as the icosahedron conjecture [8]. Theicosahedron conjecture was already solved, and the set is the optimal 3-distance set in R [18, 19] as in Theorem 1.1 (2). The following theorem plays a key role to prove our maintheorem. Theorem 1.3.
Every -distance set in R with at least points contains an s -distance setfor some s ≤ with points. The main concept to prove Theorem 1.3 is the diameter graph [7] of a subset in R d . Thediameter graph of a set X in R d is the graph on X where two vertices are adjacent if the twovertices have the largest distance in X . The subset of X corresponding to an independence setof the diameter graph does not have the largest distance in X . Thus we can verify the existenceof an s (cid:48) -distance subset of an s -distance set X with s (cid:48) < s by the independence number of its2iameter graph. The existence of s (cid:48) -distance set is useful to determine an optimal s -distanceset in low dimensions [17, 18]. Ramsey numbers or complementary Ramsey numbers [15] arealso expected to show the existence of an s (cid:48) -distance subsets of an s -distance set.In section 2, we discuss the distances in a regular dodecahedron and we enumerate thenumber of 8-point subsets of a dodecahedron which are 3- or 4-distance. In section 3, we con-sider the independence numbers of diameter graphs and prove Theorem 1.3. The classificationof 8-point s -distance sets in R for s ≤ s -distance sets are constructed from s -colorings. In section 5, we classify 8-point 3- or 4-distance sets which may be subsets of a20-point 5-distance set in R , and prove Theorem 1.2. Let G = ( V, E ) be a simple graph, where V = V ( G ) and E = E ( G ) are the vertex set andthe edge set of G , respectively. A subset W of V ( G ) is an independent set ( resp. clique ) of G if any two vertices in W are nonadjacent ( resp. adjacent). The independence number α ( G )( resp. clique number ω ( G )) of a graph G is the maximum cardinality among the independentsets ( resp. cliques) of G . Let R i = { ( x, y ) ∈ V × V | d ( x, y ) = i } , where d is the shortest-pathdistance. The i -th distance matrix A i of G is the matrix indexed by V whose ( x, y )-entryis 1 if ( x, y ) ∈ R i , and 0 otherwise. A simple graph G is a distance-regular graph [4, 6] iffor any non-negative integers i, j, k , the number p kij = |{ z ∈ V | ( x, z ) ∈ R i , ( z, y ) ∈ R j }| isindependent of the choice of ( x, y ) ∈ R k . The algebra A spanned by { A i } over the complexnumbers is called the Bose–Mesner algebra of a distance-regular graph. There exists anotherbasis { E i } such that E i E j = δ ij E i , where δ ij is the Kronecker delta. The matrices E i arecalled primitive idempotents , and the matrices are positive semidefinite. The matrices E i can be interpreted as the Gram matrices of some spherical sets that have the structure ofthe distance-regular graph, and E i are called spherical representations of the graph. Thefollowing matrices P = ( p i ( j )) j,i for A i = (cid:88) j p i ( j ) E j ,Q = ( q i ( j )) j,i for E i = 1 | V | (cid:88) j q i ( j ) A j , are called the first and second eigenmatrices , respectively. The entries of P are the eigenvaluesof A i , and the entries of Q are the inner products of the spherical representation of E i . Thefirst row q i (0) of Q is the rank of E i , that is the dimension where the representation E i exists.Let D be the vertex set of the dodecahedron with edge length 1. The set D is a 5-distance set, and let d = 1 , d , d , d , d be the 5-distances of D with 1 = d < d < d 2. Since D contains the cube with edge length τ , the other distancesin the cube are d = √ τ and d = √ τ . We can calculate d = √ τ − τ + 1 byPythagorean theorem. Let G be the dodecahedron graph G = ( V, E ), where V = D and E = { ( x, y ) | d ( x, y ) = d } . The graph G is a distance-regular graph, and d ( x, y ) = d i if and3nly if d ( x, y ) = i for each i ∈ { , , . . . , } , where d = 0. The second eigenmatrix Q of G is Q = √ −√ − / / 31 1 1 2 / − − / − − / − / −√ √ − / / − − − . There are two representations E and E in the 3-dimensional sphere. Indeed, both E and E are the dodecahedron, and the two graphs of A and A are isomorphic. Let Φ be thefield automorphism of Q ( √ 5) such that Φ( √ 5) = −√ M = ( m ij ) with m ij ∈ Q ( √ M ) is defined by applying Φ to the entries of M ,namely ˆΦ( M ) = (Φ( m ij )). It follows thatˆΦ( E ) = 3 A + Φ( √ A + A − A − Φ( √ A − A = 3 A − √ A + A − A + √ A − A = E . A principal submatrix T of E corresponds to a subset of the dodecahedron. The matrix ˆΦ( T )is a principal submatrix of E , and ˆΦ( T ) also corresponds to a subset of the dodecahedron.The two matrices T and ˆΦ( T ) may not be isomorphic as distance sets, but the two colorings ofthem are equivalent (see Section 4 for colorings). This observation gives the following lemma. Lemma 2.1. Let X be a subset of the dodecahedron in the unit sphere S . Let M be theGram matrix of X . Let ˆΦ is the map defined as above. Then M and ˆΦ( M ) are subsets of thedodecahedron, and the two colorings of them are equivalent. Now we discuss 8-point subsets of the dodecahedron which have only 3 or 4 distances. Lemma 2.2. There exists unique -distance subset of a regular dodecahedron with pointsup to isomorphism. The subset is the cube.Proof. Let G be the dodecahedron graph with relations R i = { ( x, y ) ∈ V × V | d ( x, y ) = d i } .We define the graphs G i = ( V, R i ) and G i,j = ( V, R i ∪ R j ). If for given i , the independencenumber α ( G i,j ) is less than 8 for each j (cid:54) = i , then we should take the distance d i for a 8-pointsubset. We can determine α ( G ) = 6 and α ( G ) = 5. This implies α ( G ,i ) ≤ < α ( G ,i ) ≤ < i = 1 , , 5. Thus X has both d and d . Moreover, we can determine α ( G , ) = α ( G , ) = 7 and α ( G , ) = 8, which are calculated by a computer aid. Thereforethe distances of X are d , d and d . The set X corresponding to α ( G , ) is the cube. Lemma 2.3. There exist exactly of -distance subsets of a regular dodecahedron with points up to isomorphism.Proof. An 8-point 4-distance subset of a regular dodecahedron does not contain an antipodalpair { x, − x } , otherwise X is not 4-distance. A regular dodecahedron has only 10 antipodalpairs. We choose 8 antipodal pairs from the 10 pairs, and pick out one point from eachantipodal pair, then an 8-point 4-distance set is obtained. Every 8-point 4-distance subset ofa regular dodecahedron is obtained by this manner.First we prove that if two 8-point 4-distance sets X and Y are isomorphic, then X and Y are in the same orbit of the isometry group of a regular dodecahedron. Since the sets4igure 2: Dodecahedron graph X and Y are in the same sphere, there exists an isometry σ in the orthogonal group O (3)such that X σ = Y . This implies that ( ± X ) σ = ± Y , namely the set of 8 antipodal pairsof ± X are isomorphic to that of ± Y . Since a regular dodecahedron in a given sphere isuniquely determined after one face is fixed, if each set of 8 antipodal pairs makes a face ofthe dodecahedron, then σ becomes an isometry of the dodecahedron. In order to prove thateach set of 8 antipodal pairs makes a face of the dodecahedron, we prove that it is impossibleto break all the faces of a regular dodecahedron by removing 2 antipodal pairs. If we removean antipodal pair, then 6 faces are broken. By removing one more antipodal pair, we wouldlike to break the remaining 6 faces, but it is impossible. Therefore, each set of 8 antipodalpairs contains a face of the dodecahedron, and σ becomes an isometry of the dodecahedron.We can determine the number of the 4-distance sets up to isomorphism by Burnside’slemma. The isometry group Aut( D ) of a regular dodecahedron is a subgroup of a symmetricgroup S on the 20 vertices, which is isomorphic to A × C . The vertices are indexed asFigure 2. Let N σ denote the number of the 4-distance sets fixed by σ ∈ Aut(Γ). For each σ ∈ Aut(Γ), we determine the number N σ .The identity e fixes all the 4-distance sets, namely N e = (cid:0) (cid:1) · = 11520.The transformations that fix a face of the dodecahedron are conjugates of σ = (1 2 3 4 5)(6 7 8 9 10)(11 15 14 13 12)(16 20 19 18 17) ,σ , σ , or σ . The number of the transformations is 24. The size of a subset fixed by σ isdivisible by 5. Thus N σ = 0, and similarly N σ = 0 for any transformation σ in this case.The transformations that fix an edge of the dodecahedron are conjugates of σ = (1 2)(3 6)(5 7)(4 12)(10 11)(8 13)(9 17)(14 16)(15 18)(19 20) . The number of the transformations is 15. A 4-distance set fixed by σ contains one of { , } and { , } , one of { , } and { , } , one of { , } and { , } , and one of { , } and { , } . This implies that N σ = 2 = 16, and similarly N σ = 16 for any transformation σ inthis case.The transformations that fix a vertex of the dodecahedron are conjugates of σ = (2 5 6)(3 10 12)(4 13 7)(8 14 17)(9 18 11)(15 19 16)5r σ . The number of the transformations is 20. The size of a subset fixed by σ is congruentto 0 or 1 modulo 3. Thus N σ = 0, and similarly N σ = 0 for any transformation σ in thiscase.Let τ be the transformation such that τ ( x ) = − x for any vertex x , namely τ = (1 20)(2 19)(3 18)(4 17)(5 16)(6 15)(7 14)(8 13)(9 12)(10 11) . Clearly N τ = 0.We consider the transformations that are conjugates of τ σ = (1 19 3 17 5 20 2 18 4 16)(6 14 8 12 10 15 7 13 9 11) ,τ σ , τ σ , or τ σ . The number of the transformations is 24. The size of a subset fixed by τ σ is divisible by 10. Thus N τσ = 0, and similarly N σ = 0 for any transformation σ in this case.We consider the transformations that are conjugates of τ σ = (1 19)(2 20)(3 15)(4 9)(5 14)(6 18)(7 16)(12 17) . The number of the transformations is 15. A 4-distance set fixed by τ σ may contain one of { , } and { , } , one of { , } and { , } , one of { , } and { , } , one of { , } and { , } , one of 8 and 13, or one of 10 and 11. This implies that N τσ = 2 + (cid:0) (cid:1) = 144,and similarly N σ = 144 for any transformation σ in this case.We consider the transformations that are conjugates of τ σ = (1 20)(2 16 6 19 5 15)(3 11 12 18 10 9)(4 8 7 17 13 14)or τ σ . The number of the transformations are 20. A subset fixed by τ σ must contain − x for its point x . Thus N τσ = 0, and similarly N σ = 0 for any transformation σ in this case.By Burnside’s lemma, the number of 8-point 4-distance subsets of the dodecahedron is1 | Aut(Γ) | (cid:88) σ ∈ Aut(Γ) N σ = 1120 (1 · · · · · · · · 0) = 116 . We denote a path and a cycle with n vertices by P n and C n , respectively. We denote acomplete graph of order n by K n . For X ⊂ R d , the diameter of X is defined to be themaximum value of A ( X ). Diameters give us important information when we study distancesets especially in few dimensional space. The diameter graph DG ( X ) of X ⊂ R d is the graphwith X as its vertices and where two vertices p, q ∈ X are adjacent if d ( p, q ) is the diameterof X . Let R n be the set of the vertices of a regular n -gon. Clearly DG ( R n +1 ) = C n +1 and DG ( R n ) = n · P . Note that if the independence number α ( DG ( X )) = n (cid:48) for an s -distanceset X , then the subset of X corresponding to an independence set of order n (cid:48) is an s (cid:48) -distanceset for some s (cid:48) < s .For diameter graphs for R , we have the following propositions [17]. Proposition 3.1. Let G = DG ( X ) for X ⊂ R . Then (1) G contains no C k for any k ≥ ; if G contains C k +1 , then any two vertices in V ( G ) \ V ( C k +1 ) are not adjacent andevery vertex not in the cycle is adjacent to at most one vertex of the cycle.Moreover, G contains at most one cycle. Proposition 3.2. Let G = DG ( X ) be the diameter graph of X ⊂ R with | X | = n . If G (cid:54) = C n , then we have α ( G ) ≥ (cid:100) n (cid:101) . Propositions 3.1 and 3.2 are implied from the fact that two segments with the diametermust cross if they do not share an end point.For the diameter graphs of sets in R , Dol’nikov [7] proved the following theorem. Thistheorem plays a key role of the proof of Theorem 1.3. Theorem 3.3 (Dol’nikov) . Let G = DG ( X ) be the diameter graph of X ⊂ R . If G containstwo cycles with odd lengths, then they have a common vertex. In particular, we have the following corollary. Corollary 3.4. Let G = DG ( X ) be the diameter graph of X ⊂ R with | X | = n . If G contains an odd cycle C with length m , then α ( G ) ≥ (cid:100) n − m (cid:101) .Proof. If we remove the odd cycle C from G , then any odd cycle in G is broken by Theorem 3.3.This implies G − C is a bipartite graph. Therefore we have α ( G ) ≥ α ( G − C ) ≥ (cid:100) n − m (cid:101) .In the remaining of this section, we give a proof of Theorem 1.3.By Corollary 3.4, if the diameter graph G = DG ( X ) of X ⊂ R with 20 points contains a3-cycle or a 5-cycle, then α ( G ) ≥ 8. Let G n be the set of all graphs of order n which do notcontain neither a 3-cycle nor a 5-cycle. We define f ( n ) = min { α ( G ) | G ∈ G n } . Since α ( C n ) = (cid:100) n/ (cid:101) , we have f ( n ) ≤ (cid:100) n/ (cid:101) . For a group G and S ⊂ G , we define the Cayley graph Cay( G, S ) as the graph whose vertex set is G and two vertices v, w ∈ G areadjacent if v − w ∈ S . It is easy to see that Cay( Z , {± , ± } ) does not contain neither a3-cycle nor a 5-cycle and α (Cay( Z , {± , ± } )) = 7. This implies f (17) ≤ Lemma 3.5. Let f ( n ) be defined as above. Then ≤ f ( n + 1) − f ( n ) ≤ holds.Proof. Let G ∈ G n be a graph satisfying α ( G ) = f ( n ) and G (cid:48) be the graph given by addingone isolated vertex to G . Then f ( n + 1) ≤ α ( G (cid:48) ) = α ( G ) + 1 ≤ f ( n ) + 1. Let G ∈ G n +1 be agraph satisfying α ( G ) = f ( n + 1) and H be an independent set of G with | H | = f ( n + 1). Let v ∈ V ( G ) \ H . Then H is an independent set of G − { v } , and α ( G − { v } ) = | H | . Therefore f ( n ) ≤ α ( G − { v } ) = | H | = f ( n + 1).For a vertex v ∈ V ( G ), Γ i ( v ) = { w ∈ V ( G ) | d ( v, w ) = i } , where d ( v, w ) is the shortest-path distance between v and w . We abbreviate Γ( v ) = Γ ( v ). We define G i ( v ) as theinduced subgraph with respect to Γ i ( v ) and k i ( v ) = | Γ i ( v ) | . Let m be a positive integer. Wedenote Γ ∗ m ( v ) = (cid:83) i ≥ m Γ i ( v ) and k ∗ m ( v ) = | Γ ∗ m ( v ) | . Moreover, we define G ∗ m ( v ) as the inducedsubgraph of G with respect to Γ ∗ m ( v ). Note that we regard d ( v, w ) = ∞ and w ∈ Γ ∗ m ( v ) ifthere is no path between v and w . Then the following degree condition holds.7 emma 3.6. Let n and t be positive integers. Let G ∈ G n and v ∈ V ( G ) . If α ( G ) < t Since G ∈ G n , { v } ∪ Γ ( v ) is an independent set of G . In particular, we have k ∗ ( v ) ≥ n − k ( v ) − t + 1 since 1 + k ( v ) ≤ α ( G ) < t and k ( v ) ≤ t − 2. Then t > α ( G ) ≥ k ( v ) + f ( k ∗ ( v )) ≥ k ( v ) + f ( n − k ( v ) − t + 1) , since w and w are not adjacent for any w ∈ Γ ( v ) and w ∈ Γ ∗ ( v ).For a small integer n , we can determine f ( n ) by using Lemma 3.6 Lemma 3.7. We have f (3) = 2 , f (5) = 3 , f (8) = 4 , f (10) = 5 , f (13) = 6 and f (17) = 7 .Proof. Since P , P , C , C , C and Cay( Z , {± , ± } ) are examples whose independencenumbers are the values t in the assertion. This implies the inequalities f ( n ) ≤ t for each case.It is enough to prove the converse inequalities f ( n ) ≥ t . We only prove f (17) ≥ G ∈ G suchthat α ( G ) < 7. If there exists v ∈ V ( G ) such that k ( v ) = 3, then 7 > f (8) = 3 + 4from Lemma 3.6 with t = 7, which is a contradiction. It is easy to see that we conclude acontradiction for k ( v ) > k ( v )+ f ( n − k ( v ) − t +1) ≤ ( k ( v )+1)+ f ( n − ( k ( v )+1) − t +1)holds in general from f ( n + 1) − f ( n ) ≤ k ( v ) ≤ v ∈ V ( G ). Then G is the union of cycles, paths or isolated vertices. Except for an oddcycle, each connected component of G with m vertices has an independent set of size (cid:100) m/ (cid:101) .Moreover, G contains at most one odd cycle by Theorem 3.3. Then α ( G ) ≥ 8, which is acontradiction. Therefore f (17) ≥ f (17) = 7.We can determine other f ( n ) for small n . For example we have f (12) = 5 and f (16) =7. For f (12) = 5, it is clear because 5 = f (10) ≤ f (12) and Cay( Z , {± , } ) has theindependence number 5. For f (16) = 7, it is proved by a similar way to the proof of f (17) = 7,and an attaining graph is obtained by removing one vertex from Cay( Z , {± , ± } ) whilemaintaining the independence number. However, the values in Lemma 3.7 are enough toprove Lemmas 3.8 and 3.9. Lemma 3.8. Let G be the diameter graph of X ⊂ R with | X | = 20 . If G is disconnected,then α ( G ) ≥ .Proof. Since G is disconnected, there exists a partition V = V ∪ V such that v and v arenot adjacent for any v ∈ V and v ∈ V . We may assume | V | ≥ 10. Let n i = | V i | and H i bethe induced subgraph of G with respect to V i for i = 1 , 2. Note that we may assume that both H and H do not contain neither a 3-cycle nor a 5-cycle by Corollary 3.4. If 10 ≤ n ≤ α ( H ) ≥ f ( n ) ≥ f (10) ≥ α ( H ) ≥ f ( n ) ≥ f (5) ≥ α ( G ) = α ( H ) + α ( H ) ≥ . If n = 16, then α ( G ) ≥ f (16) + f (4) ≥ f (15) + f (3) =6 + 2 = 8 since | H | ≥ 16 and | H | ≥ 4. If 17 ≤ n ≤ 19, then α ( G ) ≥ f (17) + f (1) ≥ | H | ≥ 17. Therefore α ( G ) ≥ Lemma 3.9. Let G be the diameter graph of X ⊂ R with | X | = 20 . Then α ( G ) ≥ . roof. Suppose that G contains a 3-cycle or a 5-cycle. Then α ( G ) ≥ G does not contain neither a 3-cycle nor a 5-cycle. Let v ∈ V ( G ). Since G does not contain neither a 3-cycle nor a 5-cycle, both Γ ( v ) and Γ ( v ) ∪{ v } are independent sets. Therefore we may assume k ( v ) ≤ k ( v ) ≤ 6. In particular, wemay assume k ∗ ( v ) = 20 − (1 + k ( v ) + k ( v )) ≥ − k ( v ). Moreover, we may assume that G is connected by Lemma 3.8.Suppose 5 ≤ k ( v ) ≤ v ∈ V ( G ). Since k ∗ ( v ) ≥ − k ( v ) ≥ 6, we have α ( G ∗ ( v )) ≥ f (6) ≥ f (5) = 3. Let H be an independent set of G ∗ ( v ) with | H | = 3. ThenΓ ( v ) ∪ H is an independent set of G . Therefore we have α ( G ) ≥ k ( v ) + f (5) ≥ k ( v ) = 4 for some v ∈ V ( G ). Since k ∗ ( v ) = 13 − k ( v ) ≥ 9, we have α ( G ) ≥ k ( v ) + f (9) ≥ k ( v ) + f (8) = 4 + 4 = 8. Suppose k ( v ) = 3 for some v ∈ V ( G ). Then we have k ∗ ( v ) = 13 − k ( v ) ≥ 10. Therefore α ( G ) ≥ k ( v ) + f (10) = 3 + 5 = 8. Suppose k ( w ) ≤ w ∈ V ( G ). Since k ( w ) ≤ G is connected, G is isomorphic to C or P . Thenwe have α ( G ) = 10. This completes the proof.By Lemma 3.9, we have Theorem 1.3. Let X = { p , p , . . . , p n } be an s -distance set with A ( X ) = { α , α , . . . , α s } . Let [ n ] = { , , . . . , n } and (cid:0) Sk (cid:1) = { T ⊂ S | | T | = k } for a finite set S . An s -distance set with n points isrepresented by an edge coloring of the complete graph K n by s colors. We regard an s -coloringof the edge set of K n by a surjection c : (cid:0) [ n ]2 (cid:1) → [ s ]. We define an s -coloring c : (cid:0) [ n ]2 (cid:1) → [ s ] of an s -distance set X by a natural manner, namely, c ( { i, j } ) = k where d ( p i , p j ) = α k . Conversely,an s -distance set X is called a realization of c if c is a coloring of X .Two s -colorings c and c are said to be equivalent if there exists bijections g : [ s ] → [ s ]and h : [ n ] → [ n ] such that g ( c ( { h ( i ) , h ( j ) } )) = c ( { i, j } ) for each { i, j } ∈ (cid:0) [ n ]2 (cid:1) . We definethe coloring matrix C = C ( x , x , . . . , x s ) of the coloring c with respect to x , x , . . . , x s by C i,j = (cid:40) i = j,x c ( { i,j } ) if i (cid:54) = j. In particular, C = C (1 , , . . . , s ) is called a normal coloring matrix . A coloring c is oftenrepresented as its normal coloring matrix C in this paper. We distinguish them by lowercaseletter c and uppercase letter C .For a subset X = { p , p , . . . , p n } ⊂ R d , we define the squared distance matrix D = D ( X )of X by D = ( d ( p i , p j ) ) ≤ i,j ≤ n . For an n × n symmetric matrix M = ( m i,j ) ≤ i,j ≤ n , Gram( M ) is defined to be the ( n − × ( n − i, j ) entries m i,n + m j,n − m i,j . For X = { p , p , . . . , p n } , the matrix Gram( D ( X )) is the Gram matrix of X when p n is locatedat the origin. 9 heorem 4.1. Let M be an ( n − × ( n − real symmetric matrix. There exists X in R d such that M is equal to the Gram matrix of X if and only if M = ( m i,j ) ≤ i,j ≤ n − satisfies M is positive semidefinite , rank M ≤ d,m i,i > i ∈ [ n − 1] and m i,j < m i,i + m j,j for every 1 ≤ i < j ≤ n − . (4.1)An s -coloring c : (cid:0) [ n ]2 (cid:1) → [ s ] is said to be representable in R d if there exists distinct realnumbers α , α , . . . , α s such that C ( α , α , . . . , α s ) satisfies (4.1). To decide an s -coloring arenot representable, the rank condition in (4.1) is effective. An s -coloring c : (cid:0) [ n ]2 (cid:1) → [ s ] is saidto be quasi representable in R d if there exists distinct complex numbers α , α , . . . , α s suchthat rank C ( α , α , . . . , α s ) ≤ d .For a square matrix M = ( m i,j ) ≤ i,j ≤ n and an index set T = { t , t , . . . , t k } ∈ (cid:0) [ n ] k (cid:1) , wedefine a principal submatrix of M with respect to T bysub( M ; T ) = ( m t i ,t j ) ≤ i,j ≤ k . Let M k ( M ) = (cid:26) sub( M ; T ) | T ∈ (cid:18) [ n ] k (cid:19)(cid:27) . We define r ( M ) = max { k ∈ [ n ] | ∃ S ∈ M k , det S (cid:54) = 0 } . Proposition 4.2. For a square matrix M , r ( M ) ≤ rank M holds. Moreover, r ( M ) = rank M holds if M is positive semidefinite.Proof. It is well known that rank M is the maximum value k such that there exists a squaresubmatrix S of size k in M with det S (cid:54) = 0 that may not be principal. This implies r ( M ) ≤ rank M . Suppose M is a positive semidefinite matrix of size n . Since M is positivesemidefinite, there exists n × rank M matrix N such that M = N N (cid:62) and rank N = rank M .For T ∈ (cid:0) [ n ] k (cid:1) , let χ T be the n × n diagonal matrix with diagonal entries ( χ T ) ii = 1 if i ∈ T , and ( χ T ) ii = 0 if i (cid:54)∈ T . For a row vector x ∈ R n and T ∈ (cid:0) [ n ] k (cid:1) , it follows that x ( χ T N N (cid:62) χ (cid:62) T ) x (cid:62) = 0 if and only if x ( χ T N ) = 0. Thus,rank( χ T N N (cid:62) χ (cid:62) T ) = n − dim { x ∈ R n | x ( χ T N N (cid:62) χ (cid:62) T ) x (cid:62) = 0 } = n − dim { x ∈ R n | x ( χ T N ) = 0 } = rank( χ T N ) . For T ∈ (cid:0) [ n ] k (cid:1) , it follows that det(sub( M ; T )) (cid:54) = 0 if and only if rank( χ T N N (cid:62) χ (cid:62) T ) ≥ k . Thisimplies that r ( M ) is the maximum value k such that rank( χ T N ) = k , which is k = rank N =rank M .An s -coloring c : (cid:0) [ n ]2 (cid:1) → [ s ] is said to be weakly quasi representable in R d if there existdistinct complex numbers α , α , . . . , α s such that r ( C ( α , α , . . . , α s )) ≤ d . Clearly if an s -coloring c is representable in R , then c is a (weakly) quasi representable in R . Thefollowing proposition is essentially proved by Sz¨oll˝osi and ¨Osterg˚ard [19], but we should takeall submatrices M that may not be principal in their result. Actually, it is enough to use allprincipal submatrices M to collect our desired colorings.10 (cid:93) QRC 512 62095 4499 1093 277 59 12 5 2 0Table 2: Number of quasi representable at most 4-colorings in R of n points Proposition 4.3. An s -coloring c : (cid:0) [ n ]2 (cid:1) → [ s ] is a weakly quasi representable in R if andonly if the following system of equations in s values (cid:40) det M = 0 for all M ∈ M (Gram( C (1 , x , x , . . . , x s − ))) , u (cid:81) s − i =1 x i ( x i − (cid:81) ≤ j 4. We will prove the following theoremin this section. Theorem 5.1. Let Y be an s -distance set in R with points for ≤ s ≤ . If Y ∪ Z is a -distance set in R with at least points, then Y ∪ Z is isomorphic to a regular dodecahedron. Note that there exists no 2-distance set with 8 points in R . In this section, firstly, weconsider (weak) quasi representable s -colorings c in R instead of s -distance sets in R . Thenwe consider realizations of c as needed.Sz¨oll˝osi and ¨Osterg˚ard [19] classified quasi representable s -colorings in R for s ≤ 4, seeTable 2. Lemma 5.2 (Sz¨oll˝osi and ¨Osterg˚ard [19]) . There exist exactly quasi representable -colorings in R with vertices and exactly quasi representable -colorings in R with vertices. We denote the set of all quasi representable s -colorings in R with n vertices by CG ( n, s ).By Lemma 5.2, we have |CG (8 , | = 19 and |CG (8 , | = 1074.Let C ∈ CG (8 , ∪CG (8 , G ( C ) = ( V, E ) with respect to C as follows.By a computer search, we find all vectors ( a , a , . . . , a ) ∈ [5] such that M = a · · · a a ... Ca (5.1)is a weakly quasi representable s -colorings in R for s ≤ M is weakly quasi representable, first we calculate a Gr¨obner basis B ofsystem (4.2) for C , see [19] about the manner in details. Then, we calculate a Gr¨obner basisfor the union of B and the set of the first equations in (4.2) for all sub( M ; T ) with 1 ∈ T ,which determine whether M is weakly quasi representable. Throughout this paper, computercalculations are done with functions of the software Magma [3] and Maple [14].11e regard the set of all vectors satisfying (5.1) as the vertex set V of the graph G ( C ).Then two vertices ( a , a , . . . , a ) , ( a (cid:48) , a (cid:48) , . . . , a (cid:48) ) ∈ V are adjacent if there exists i ∈ [5] suchthat i a · · · a i a (cid:48) · · · a (cid:48) a a (cid:48) ... ... Ca a (cid:48) (5.2)is a weakly quasi representable s -coloring in R for s ≤ 5. Some special graphs haveloops as Lemma 5.3 below. For positive real numbers α , α , . . . , α n and a subset X = { p , p , . . . , p n } ⊂ R which is not co-linear, there exist at most two points q ∈ R suchthat d ( p i , q ) = α i for any i ∈ [ n ]. In particular, if there exist two points which satisfy thiscondition, then X is co-planar. By exhaustive computer search, we have the following lemma. Lemma 5.3. Let C ∈ CG (8 , ∪ CG (8 , . G ( C ) has a loop if and only if a realization of C is isomorphic to one of the following nine -distance sets. (a) the subset with points of a regular nonagon, (b) a regular octagon, (c) the six subsets with points of the set in Figure 1 (a) in Theorem 1.1, (d) the set in Figure 1 (d) in Theorem 1.1.Moreover, there exist two loops only for (d) and there is only one loop for other cases. For C ∈ CG (8 , ∪ CG (8 , ω ∗ ( C ) = ω ( G ( C )) + l ( G ( C )), where ω ( G ( C )) is the cliquenumber of G ( C ) and l ( G ( C )) is the number of loops in G ( C ). To prove Theorem 5.1, it isenough to classify C ∈ CG (8 , ∪ CG (8 , 4) such that ω ∗ ( C ) ≥ 12. By exhaustive computersearch, we have the following lemma. Lemma 5.4. (i) There exists a unique coloring C ∈ CG (8 , with ω ∗ ( C ) ≥ , whichcorresponds to the cube. Moreover, ω ∗ ( C ) = ω ( G ( C )) = 12 and there exists the uniqueclique of order for the coloring. (ii) There exist exactly colorings C ∈ CG (8 , with ω ∗ ( C ) ≥ . Moreover, ω ∗ ( C ) = ω ( G ( C )) = 12 and there exists the unique clique of order for each coloring amongthe colorings. We classify 8-point s -distance sets for s ≤ s -colorings in Lemma 5.4. If a realization X of a coloring C is a subset of the dodecahedron,then we have another realization ˆΦ( X ) of C by Lemma 2.1. The realizations X and ˆΦ( X )may be isomorphic.Let C ∈ CG (8 , 3) be the coloring in Lemma 5.4 (i) and X be a realization of C . Then wehave A ( X ) = { , √ , √ } by solving system (4.2). Then it is easy to see that X is a cube.12et C = . Then C ∈ CG (8 , 4) is a coloring in Lemma 5.4 (ii). There exist four solutions of system (4.2)for C and there exist exactly four realizations of C up to isomorphism. Let W = { , , , , , , , } (5.3)be a subset of the vertex set of the dodecahedron graph as given in Figure 2. The shortest-pathdistance matrix ( d ( x, y )) x,y ∈ W of W is C , where C is indexed by W using the order of ele-ments in (5.3). Four realizations of C are given in Figure 3. Let Y = { A , A , . . . , A } , Y = { B , B , . . . , B } and X i = Y i ∪ { P } , X (cid:48) i = Y i ∪ { P (cid:48) } ( i = 1 , , where P (cid:48) i is the reflection of P i in the plane π i for i = 1 , 2. Then X , X , X (cid:48) and X (cid:48) are all therealizations of C . Both X and X are subsets of the dodecahedron and have the structureof the coloring C , that shows the situation of Lemma 2.1. There exist exactly two solutionsof system (4.2) for the other 62 colorings C ∈ CG (8 , C ∈ CG (8 , 4) is one of the ten colorings obtained from { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , { , , , , , , , } , by the above manner, then the two realizations of C are isomorphic. Except for the above10 colorings and C , each C ∈ CG (8 , 4) in Lemma 5.4 (ii) has exactly two realizations up toisomorphism. Then we have the following lemma. Lemma 5.5. (i) There exists a unique -distance set whose coloring is given in Lemma 5.4 (i). (ii) Among the colorings in Lemma 5.4 (ii), we have the following: (a) There exists exactly one coloring which has four solutions of system (4.2) and thefour realizations corresponding to the solutions are not isomorphic to each other. (b) There exist exactly colorings which have two solutions of system (4.2) and thetwo realizations corresponding to the solutions are not isomorphic. (c) There exist exactly colorings which have two solutions of system (4.2) but thetwo realizations corresponding to the solutions are isomorphic. C By Lemma 5.5, there exist exactly 118 of 4-distance sets in R given from the colorings inLemma 5.4. The two sets X (cid:48) and X (cid:48) are not subsets of the dodecahedron. By Lemma 2.3,the remaining 116 4-distance sets should be subsets of the dodecahedron. Note that the cubeis also a subset of the dodecahedron. Let S be the set of the cube and the 116 4-distancesubsets. By Lemma 5.4, ω ( G ( C )) = 12 for the coloring C obtained from X ∈ S , and thecorresponding clique of order 12 is unique. Therefore a 20-point 5-distance set that contains X ∈ S must be the dodecahedron.In order to prove Theorem 5.1, we prove that for i = 1 , Z ⊂ R such that X (cid:48) i ∪ Z is a 20-point 5-distance set. We consider a candidate P ∈ R such that | A ( X (cid:48) i ∪ { P } ) | ≤ i = 1 , { Q (cid:48) , R (cid:48) , O } for X (cid:48) and { Q (cid:48) , R (cid:48) , O } for X (cid:48) in Figure 3. The points Q (cid:48) i and R (cid:48) i are the reflections of Q i and R i inthe plane π i , respectively. Moreover, O ( resp. O (cid:48) ) is the center of the pentagon consistedof { A , A , . . . , A } ( resp. { B , B , . . . , B } ). Thus the cardinality of a 5-distance set thatcontains X (cid:48) i is at most 11. Therefore a proof of Theorem 5.1 is complete.Finally, we prove Theorem 1.2. Proof of Theorem 1.2. 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(4) (2012), iroshi Nozaki Department of Mathematics Education, Aichi University of Education, 1 Hirosawa, Igaya-cho, Kariya, Aichi 448-8542, Japan.E-mail address: [email protected] Masashi Shinohara Faculty of Education, Shiga University, 2-5-1 Hiratsu, Otsu, Shiga 520-0862, Japan.E-mail address: [email protected]@edu.shiga-u.ac.jp